When analysing the outcomes and probabilities of a Discrete Random Variable, we become interested in the expected value of the distribution.
Based on the probabilities, the expected value is which outcome we expect to occur within the given experiment.
The expected value, denoted $E(X)$E(X), can also be thought of as the mean of the distribution, and is in fact calculated like a weighted average.
Consider the probability distribution of a DRV given in the table below.
$x$x | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
---|---|---|---|---|---|
$P(X=x)$P(X=x) | $0.2$0.2 | $0.15$0.15 | $0.4$0.4 | $0.1$0.1 | $0.15$0.15 |
To calculate the mean, we multiply each outcome by its probability. The probability represents the weighting for each outcome occurring.
$E\left(x\right)=1\times0.2+2\times0.15+3\times0.4+4\times0.1+5\times0.15$E(x)=1×0.2+2×0.15+3×0.4+4×0.1+5×0.15
$E\left(x\right)=2.85$E(x)=2.85
Let's consider what this means. Clearly we're not really "expecting" an outcome of $2.85$2.85. All possible outcomes are discrete and are listed in the table.
It's just like when they say the average number of children per family is $2.3$2.3. No family will have $0.3$0.3 of a child! But the mean value gives us an idea about how many children we can expect to see in a family; somewhere between two and three.
So in our experiment we don't expect to get $2.85$2.85, but in the long term we expect the average of all our occurrences to come to $2.85$2.85.
Similarly, the expected value is not to be confused with the most likely outcome. In our distribution above, the most likely outcome is $3$3 since $P(X=3)$P(X=3) has the highest probability of $0.4$0.4.
Let's take a look at one more example to see another way using the expected value can crop up.
Consider the probability distribution of a DRV given in the table below, where $E(X)=3.05$E(X)=3.05
$x$x | $0$0 | $1$1 | $3$3 | $5$5 | $6$6 |
---|---|---|---|---|---|
$P(X=x)$P(X=x) | $0.3$0.3 | $a$a | $0.1$0.1 | $b$b | $0.15$0.15 |
Determine the values of $a$a and $b$b.
Since we have two unknowns we need two equations to use in solving for $a$a and $b$b.
We are told that the table represents a DRV, so we know all the probabilities must add to $1$1. So our first equation is:
$0.3+a+0.1+b+0.15=1$0.3+a+0.1+b+0.15=1
Simplifying gives us:
$a+b=0.45$a+b=0.45
Using the expected value we can form another equation.
$0\times0.3+1\times a+3\times0.1+5\times b+6\times0.15=3.05$0×0.3+1×a+3×0.1+5×b+6×0.15=3.05
Simplifying gives us:
$a+5b=1.85$a+5b=1.85
Solving using the elimination method or using the solving functionality of your calculator gives us:
$4b=1.4$4b=1.4
$b=0.35$b=0.35 and $a=0.1$a=0.1
Consider the probability density function defined by:
$P$P$($($X=x$X=x$)$) | $=$= | $\frac{x}{25}$x25; $x=3$x=3, $4$4, $5$5, $6$6, $7$7 | |
$0$0, otherwise |
Does this probability density function represent the distribution for a discrete random variable?
Yes
No
Yes
No
Hence complete the table of values for this probability distribution.
$x$x | $3$3 | $4$4 | $5$5 | $6$6 | $7$7 |
---|---|---|---|---|---|
$P$P$($($X=x$X=x$)$) | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Now calculate the expected value of the distribution.
Examine the graph of the probability distribution below.
Calculate the expected value of the discrete probability distribution.
Is the median of this distribution higher, lower or equal to the mean?
equal
higher
lower
equal
higher
lower
What is the reason for your answer in part b of this question?
The graph is negatively skewed.
The graph is positively skewed.
The graph is symmetrical.
The graph is negatively skewed.
The graph is positively skewed.
The graph is symmetrical.
Investigate situations that involve elements of chance: A calculating probabilities of independent, combined, and conditional events B calculating and interpreting expected values and standard deviations of discrete random variables C applying distributions such as the Poisson, binomial, and normal
Apply probability distributions in solving problems